标签:hint fir nta ref als you number upd script
http://poj.org/problem?id=3468
You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum of numbers in a given interval.
The first line contains two numbers N and Q. 1 ≤ N,Q ≤ 100000.
The second line contains N numbers, the initial values of A1, A2, ... , AN. -1000000000 ≤ Ai ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of Aa, Aa+1, ... , Ab. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of Aa, Aa+1, ... , Ab.
You need to answer all Q commands in order. One answer in a line.
10 5 1 2 3 4 5 6 7 8 9 10 Q 4 4 Q 1 10 Q 2 4 C 3 6 3 Q 2 4
4 55 9 15
1 #include <stdio.h> 2 #include <string.h> 3 #include <iostream> 4 #include <string> 5 #include <math.h> 6 #include <algorithm> 7 #include <vector> 8 #include <queue> 9 #include <set> 10 #include <map> 11 #include <math.h> 12 const int INF=0x3f3f3f3f; 13 typedef long long LL; 14 const int mod=1e9+7; 15 //const double PI=acos(-1); 16 const int maxn=1e5+10; 17 using namespace std; 18 //ios::sync_with_stdio(false); 19 // cin.tie(NULL); 20 21 int n,m; 22 struct node 23 { 24 int l; 25 int r; 26 LL lazy;//注意lazy也要开LL 27 LL sum; 28 }SegTree[maxn<<2]; 29 30 void PushUp(int rt) 31 { 32 SegTree[rt].sum=SegTree[rt<<1].sum+SegTree[rt<<1|1].sum; 33 } 34 35 void PushDown(int rt) 36 { 37 if(SegTree[rt].lazy!=0) 38 { 39 SegTree[rt<<1].lazy+=SegTree[rt].lazy; 40 SegTree[rt<<1|1].lazy+=SegTree[rt].lazy; 41 SegTree[rt<<1].sum+=(SegTree[rt<<1].r-SegTree[rt<<1].l+1)*SegTree[rt].lazy; 42 SegTree[rt<<1|1].sum+=(SegTree[rt<<1|1].r-SegTree[rt<<1|1].l+1)*SegTree[rt].lazy; 43 SegTree[rt].lazy=0; 44 } 45 } 46 47 void Build(int l,int r,int rt) 48 { 49 SegTree[rt].l=l; 50 SegTree[rt].r=r; 51 SegTree[rt].lazy=0;//多样例时必须加 52 if(l==r) 53 { 54 scanf("%lld",&SegTree[rt].sum); 55 return; 56 } 57 int mid=(l+r)>>1; 58 Build(l,mid,rt<<1); 59 Build(mid+1,r,rt<<1|1); 60 PushUp(rt); 61 } 62 63 void Update(int L,int R,int add,int rt) 64 { 65 int l=SegTree[rt].l; 66 int r=SegTree[rt].r; 67 if(L<=l&&R>=r) 68 { 69 SegTree[rt].sum+=(SegTree[rt].r-SegTree[rt].l+1)*add; 70 SegTree[rt].lazy+=add; 71 return ; 72 } 73 PushDown(rt);//向下更新lazy 74 int mid=(l+r)>>1; 75 if(L<=mid) 76 Update(L,R,add,rt<<1); 77 if(R>mid) 78 Update(L,R,add,rt<<1|1); 79 PushUp(rt); 80 } 81 82 LL Query(int L,int R,int rt) 83 { 84 int l=SegTree[rt].l; 85 int r=SegTree[rt].r; 86 if(L<=l&&R>=r) 87 { 88 return SegTree[rt].sum; 89 } 90 PushDown(rt);//向下更新lazy 91 int mid=(l+r)>>1; 92 LL sum=0; 93 if(L<=mid) 94 sum+=Query(L,R,rt<<1); 95 if(R>mid) 96 sum+=Query(L,R,rt<<1|1); 97 return sum; 98 } 99 100 int main() 101 { 102 while(~scanf("%d %d",&n,&m)) 103 { 104 Build(1,n,1); 105 for(int i=1;i<=m;i++) 106 { 107 char c[5]; 108 int a,b; 109 scanf("%s %d %d",c,&a,&b); 110 if(c[0]==‘Q‘) 111 { 112 printf("%lld\n",Query(a,b,1)); 113 } 114 else if(c[0]==‘C‘) 115 { 116 int add; 117 scanf("%d",&add); 118 Update(a,b,add,1); 119 } 120 } 121 } 122 return 0; 123 }
POJ-3468 - A Simple Problem with Integers(线段树区间更新模板)
标签:hint fir nta ref als you number upd script
原文地址:https://www.cnblogs.com/jiamian/p/11392142.html